Let $(W_t), 0\leq t \leq T $ be a wiener process.
a) $\mathbb{E}[W_T^2|W_\frac{3T}{4}=0]=?$
My guess is that the solution should be $\frac{9T^2}{16}$ because the expected value of $(W_T^2)$ would be $t$ but the process starts at $\frac{3T}{4}$
b) $\mathbb{P}(W_T - T < 0 | W_\frac{3T}{4} = T ) = ?$
Here I think that the solution should be the distribution function of the normal distribution:
$\mathbb{P}(W_T - T < 0 | W_\frac{3T}{4} = T ) = \mathbb{P}(W_T - W_\frac{3T}{4} < 0 | W_\frac{3T}{4} = T ) = \mathbb{P}(W_T < W_\frac{3T}{4} | W_\frac{3T}{4} = T )= \int_{-{\infty}}^\frac{3T}{4}\frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}(\frac{x^2}{T-\frac{3T}{4}})}dx$